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Homework Chapter 7
1. A manufacturer claims that the average life of its light bulb is 1,200 hours. Assume that the
life of all such bulbs has a normal distribution with mean 1,200 hours, and standard deviation
70 hours. Determine the probability the mean life time of a random sample of 9 bulbs will be:
a) At least 1,180 hours.
b) Between 1,185 and 1,190 hours
Solution:
µ = 1200 and σ = 70 and n = 9
a. Since the population normal also the sample mean normal with µx = 1200 and σx = 23.33
P(Z > - 0.85) = 1 – P(Z < - 0.85) = 1 – 0.1977 = 0.8023
b. P(- 0.64 < Z < - 0.43) = P(Z < -0.43) – P(Z < - 0.64) = 0.3336 – 0.2611 = 0.0725
2. A manufacturer claims that the average life of its light bulb is 1,200 hours. Assume that the
life of all such bulbs with mean 1,200 hours, and standard deviation 70 hours. Determine the
probability the mean life time of a random sample of 49 bulbs will be:
a) At least 1,170 hours.
b) Between 1,180 and 1,195 hours
Solution
Since n large (n = 49) by C.L.T the sample mean approximately normal with µx = 1200 and
σx = 10
a. P(Z > - 3) = 1 – P(Z < - 3) = 1 – 0.0013 = 0.9987
b. P(- 2 < Z < - 0.5) = P(Z < -0.5 )- P(Z < -2) = 0.3085 - 0.2275 = 0.081
3. A research indicates that 42% of all customers of Prince Burger are teenagers.
What is the probability that a random sample of 225 customers of this fast food chain would
contain more than 45% teenagers?
Solution:
π = 0.42 and n = 225
since n π =(225)( 0.42) = 94.5 and n (1- π) =(225)( 1- 0.42) = 130.5
p approximately normal with µp = 0.42 and σx = 0.032904
P(p > 0.45) = P(Z > 0.91) = 1 – 0.8186 = 0.1814
4. A recent study in Riyadh has concluded that the distance that workers daily travel to work
averages 19.3 km with a standard deviation of 5.6 km. Assuming these values correspond to
the parameters of a population,
a) What is the probability that a sample of 49 yields a mean less than 20 km?
b) What is the value of the sample mean such that 97% of the mean above this value?
Solution:
a. Since n large (n = 49), by C.L.T the sample mean approximately normal with µx =
19.3 and σx = 0.8
P(Z < 0.88) = 0.8106
b. P(Z < b) 0.97
b = 0.25
the sample mean = 0.25( 0.8) + 19.3 = 19.5
5. The proportion of customers who pay their shopping at a supermarket with a credit card is
0.35. If 90 customers are randomly selected, what is the probability that at least 30 paid their
bill with a credit card?
Solution:
π = 0.35 and n = 90
since n π =(90)( 0.35) = 31.5 and n (1- π) =(90)( 1- 0.35) = 58.5
with µx = 31.5 and σx = 5.524931
P(X ≥ 30) = P(X > 30.5) = P(Z > - 0.22) = 1 – P(Z < - 0.22) = 1- 0.4130 = 0.587